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If the lines 2x-3y=5 and 3x-4y=7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.A. `x^(2)+y^(2)+2x-2y=62`B. `x^(2)+y^(2)+2x-2y=47`C. `x^(2)+y^(2)-2x+2y=47`D. `x^(2)+y^(2)-2x+2y=62` |
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Answer» Correct Answer - C Since, 2x-3y = 5 and 3x-4y = 7 are diameters of a circle. Their point of intersection is centre (1, -1) Are given, `pir^(2) = 154` `rArr r^(2) = 154 xx (7)/(22) rArr r = 7 ` ` therefore` Required of circle is `(x-1)^(2) + (y+1)^(2) = 7^(2)` `rArr x^(2) + y^(2)-2x +2y = 47` |
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